∫_( 0) ^( 3) (dx/(2 + cosx)) = ?


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Question Number 147626 by mathdanisur last updated on 22/Jul/21

$\underset{0}{\int^{3}} \frac{d x}{2 + c o s x} = ?$
	Answered by mathmax by abdo last updated on 22/Jul/21

	$I = \int_{0}^{3} \frac{dx}{2 + cosx} =_{\tan (\frac{x}{2}) = y} \int_{0}^{\tan (\frac{3}{2})} \frac{2 dy}{(1 + y^{2}) (2 + \frac{1 - y^{2}}{1 + y^{2}})}$ $= 2 \int_{0}^{\tan (\frac{3}{2})} \frac{dy}{2 + {2 y}^{2} + 1 - y^{2}} = 2 \int_{0}^{\tan (\frac{3}{2})} \frac{dy}{3 + y^{2}}$ $=_{y = \sqrt{3} z} 2 \int_{0}^{\frac{1}{\sqrt{3}} \tan (\frac{3}{2})} \frac{\sqrt{3} dz}{3 (1 + z^{2})} = \frac{2 \sqrt{3}}{3} {[arctanz]}_{0}^{\frac{1}{\sqrt{3}} \tan (\frac{3}{2})} \Rightarrow$ $I = \frac{2 \sqrt{3}}{3} \arctan (\frac{1}{\sqrt{3}} \tan (\frac{3}{2}))$
	Commented by mathdanisur last updated on 22/Jul/21

	$t h a n k y o u S i r$
	Answered by gsk2684 last updated on 22/Jul/21

	$\underset{x = 0}{\int^{3}} \frac{1}{2 + (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})} d x = \underset{0}{\int^{3}} \frac{\sec^{2} \frac{x}{2}}{3 + \tan^{2} \frac{x}{2}} d x$ $p u t \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t$ $i f x = 0 t h e n t = 0$ $i f x = 3 t h e n t = \tan \frac{3}{2}$ $= \underset{t = 0}{\int^{\tan \frac{3}{2}}} \frac{2}{{(\sqrt{3})}^{2} + t^{2}} d t = 2 {[\frac{1}{\sqrt{3}} \tan^{- 1} \frac{t}{\sqrt{3}}]}_{0}^{\tan \frac{3}{2}}$ $= \frac{2}{\sqrt{3}} [\tan^{- 1} (\frac{\tan \frac{3}{2}}{\sqrt{3}}) - \tan^{- 1} 0]$ $= \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{\tan \frac{3}{2}}{\sqrt{3}})$
	Commented by mathdanisur last updated on 22/Jul/21

	$t h a n k y o u S i r$
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